3.8 \(\int \cos ^4(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=67 \[ -\frac {i a \cos ^4(c+d x)}{4 d}+\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a x}{8} \]

[Out]

3/8*a*x-1/4*I*a*cos(d*x+c)^4/d+3/8*a*cos(d*x+c)*sin(d*x+c)/d+1/4*a*cos(d*x+c)^3*sin(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3486, 2635, 8} \[ -\frac {i a \cos ^4(c+d x)}{4 d}+\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x]),x]

[Out]

(3*a*x)/8 - ((I/4)*a*Cos[c + d*x]^4)/d + (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*Cos[c + d*x]^3*Sin[c + d*x
])/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+i a \tan (c+d x)) \, dx &=-\frac {i a \cos ^4(c+d x)}{4 d}+a \int \cos ^4(c+d x) \, dx\\ &=-\frac {i a \cos ^4(c+d x)}{4 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} (3 a) \int \cos ^2(c+d x) \, dx\\ &=-\frac {i a \cos ^4(c+d x)}{4 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{8} (3 a) \int 1 \, dx\\ &=\frac {3 a x}{8}-\frac {i a \cos ^4(c+d x)}{4 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 46, normalized size = 0.69 \[ \frac {a \left (8 \sin (2 (c+d x))+\sin (4 (c+d x))-8 i \cos ^4(c+d x)+12 c+12 d x\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x]),x]

[Out]

(a*(12*c + 12*d*x - (8*I)*Cos[c + d*x]^4 + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d)

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 56, normalized size = 0.84 \[ \frac {{\left (12 \, a d x e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 6 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/32*(12*a*d*x*e^(2*I*d*x + 2*I*c) - I*a*e^(6*I*d*x + 6*I*c) - 6*I*a*e^(4*I*d*x + 4*I*c) + 2*I*a)*e^(-2*I*d*x
- 2*I*c)/d

________________________________________________________________________________________

giac [A]  time = 0.54, size = 103, normalized size = 1.54 \[ \frac {{\left (12 \, a d x e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - i \, a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) - i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 6 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

1/32*(12*a*d*x*e^(2*I*d*x + 2*I*c) + I*a*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - I*a*e^(2*I*d*x + 2
*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) - I*a*e^(6*I*d*x + 6*I*c) - 6*I*a*e^(4*I*d*x + 4*I*c) + 2*I*a)*e^(-2*I*d*x
- 2*I*c)/d

________________________________________________________________________________________

maple [A]  time = 0.44, size = 53, normalized size = 0.79 \[ \frac {-\frac {i a \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+I*a*tan(d*x+c)),x)

[Out]

1/d*(-1/4*I*a*cos(d*x+c)^4+a*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c))

________________________________________________________________________________________

maxima [A]  time = 0.92, size = 61, normalized size = 0.91 \[ \frac {3 \, {\left (d x + c\right )} a + \frac {3 \, a \tan \left (d x + c\right )^{3} + 5 \, a \tan \left (d x + c\right ) - 2 i \, a}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/8*(3*(d*x + c)*a + (3*a*tan(d*x + c)^3 + 5*a*tan(d*x + c) - 2*I*a)/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1))/
d

________________________________________________________________________________________

mupad [B]  time = 3.34, size = 64, normalized size = 0.96 \[ \frac {3\,a\,x}{8}+\frac {\frac {3\,a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{8}+\frac {3{}\mathrm {i}\,a\,\mathrm {tan}\left (c+d\,x\right )}{8}+\frac {a}{4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^3+{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i),x)

[Out]

(3*a*x)/8 + (a/4 + (a*tan(c + d*x)*3i)/8 + (3*a*tan(c + d*x)^2)/8)/(d*(tan(c + d*x) + tan(c + d*x)^2*1i + tan(
c + d*x)^3 + 1i))

________________________________________________________________________________________

sympy [A]  time = 0.29, size = 139, normalized size = 2.07 \[ \frac {3 a x}{8} + \begin {cases} - \frac {\left (256 i a d^{2} e^{6 i c} e^{4 i d x} + 1536 i a d^{2} e^{4 i c} e^{2 i d x} - 512 i a d^{2} e^{- 2 i d x}\right ) e^{- 2 i c}}{8192 d^{3}} & \text {for}\: 8192 d^{3} e^{2 i c} \neq 0 \\x \left (- \frac {3 a}{8} + \frac {\left (a e^{6 i c} + 3 a e^{4 i c} + 3 a e^{2 i c} + a\right ) e^{- 2 i c}}{8}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c)),x)

[Out]

3*a*x/8 + Piecewise((-(256*I*a*d**2*exp(6*I*c)*exp(4*I*d*x) + 1536*I*a*d**2*exp(4*I*c)*exp(2*I*d*x) - 512*I*a*
d**2*exp(-2*I*d*x))*exp(-2*I*c)/(8192*d**3), Ne(8192*d**3*exp(2*I*c), 0)), (x*(-3*a/8 + (a*exp(6*I*c) + 3*a*ex
p(4*I*c) + 3*a*exp(2*I*c) + a)*exp(-2*I*c)/8), True))

________________________________________________________________________________________